Wave-particle duality and localization

[amjad-sh]

I have read recently that the motion of an electron of momentum p must be described by the means of a plane waves :[itex]\psi(\vec r,t)=Ae^{i(\vec k \cdot \vec r -wt)}=Ae^{i(\vec p\cdot \vec r -Et)/\hbar}[/itex]
de Broglie hypothesis states that every particle of momentum p has a wavelength lamda.

I will split my question into three parts:
My first part concerns the plane wave by itself:
1) Why plane waves are written like this [itex]\psi(\vec r,t)=Ae^{i(\vec k \cdot \vec r -wt)}[/itex]
why not like this for example:[itex]\psi(\vec r,t)=Ae^{i(\vec k \cdot \vec r +wt)}[/itex]
2) Is it called a plane wave since in the far region they will approximately be like a plane?
3)what does the imaginary part of this wave means physically?

My second part concerns the wavefunction:
1) If the electron of momentum p is described by means of a plane wave, does this mean that we can't predict at all the position of the particle? since the plane waves has no sensible normalization ( [itex]|\psi(x)|^2=A[/itex] )in all the space.
2) If the answer of the above question is yes, then why in the cases of interference and diffraction the position of the electron can be predicted as there are bright,dark and intermediate fringes.Does the wavefunction change, in this case, from a plane wave to another wavefunction that have a sensible normalization?Can we relate this to wavepackets?

My third part concerns de Broglie hypothesis:
"Whenever the de Broglie wavelength of an object is in the range of, or exceeds its size, the wave nature of the object is detectable,hence it cannot be neglected.But if de Broglie wavelength is much too small compared to its size,the wave behavior of this object is undetectable".
Can somebody give an example that show me this?
THANKS!

 

[blue_leaf77]

I have read recently that the motion of an electron of momentum p must be described by the means of a plane waves :[itex]\psi(\vec r,t)=Ae^{i(\vec k \cdot \vec r -wt)}=Ae^{i(\vec p\cdot \vec r -Et)/\hbar}[/itex]
de Broglie hypothesis states that every particle of momentum p has a wavelength lamda.

That's the wavefunction of a free electron.

I will split my question into three parts:
My first part concerns the plane wave by itself:
1) Why plane waves are written like this [itex]\psi(\vec r,t)=Ae^{i(\vec k \cdot \vec r -wt)}[/itex]
why not like this for example:[itex]\psi(\vec r,t)=Ae^{i(\vec k \cdot \vec r +wt)}[/itex]
2) Is it called a plane wave since in the far region they will approximately be like a plane?
3)what does the imaginary part of this wave means physically?

1) Because the time evolution operator looks like ##e^{-iHt/\hbar}##.
2) By saying "far", you are automatically considering something distinctive in space, from which you measure distances and say that a point is "far" if it is located a great distance from this thing. In free space (i.e no potential), obviously there is no way you can say "far" as the space is homogenous. However, in some physical applications like electron scattering off certain potential or other objects, it's usually sufficient to assume that far away from the scatterer, the electron has a plane wave wavefunction.
3) The imaginary part of a wavefunction means as significant as does the real part. The wavefunction is a complex quantity - I don't think you will get more insight by studying the real and imaginary parts separately.

 

[amjad-sh]

1) Because the time evolution operator looks like e−iHt/ℏe−iHt/ℏe^{-iHt/\hbar}.

3) The imaginary part of a wavefunction means as significant as does the real part. The wavefunction is a complex quantity - I don't think you will get more insight by studying the real and imaginary parts separately.

Ok, then the plane wave here came from the schrodinger equation.

2) By saying "far", you are automatically considering something distinctive in space, from which you measure distances and say that a point is "far" if it is located a great distance from this thing. In free space (i.e no potential), obviously there is no way you can say "far" as the space is homogenous. However, in some physical applications like electron scattering off certain potential or other objects, it's usually sufficient to assume that far away from the scatterer, the electron has a plane wave wavefunction.

Thanks, but you still didn't tell me why it is named a plane wave?

 

[blue_leaf77]

Thanks, but you still didn't tell me why it is named a plane wave?

Because the mathematical expression looks like that of a plane wave.

 

[amjad-sh]

Because the mathematical expression looks like that of a plane wave.

I mean Why Acos(kx -wt) is the expression of the plane wave?
I can't find a correlation between the "plane" and this expression.
I don't know if you got what I mean.

 

[Mentz114]

I mean Why Acos(kx -wt) is the expression of the plane wave?
I can't find a correlation between the "plane" and this expression.
I don't know if you got what I mean.

It is geometry. The direction ##\vec{k}## defines an orthogonal plane of equal momentum.

 

[amjad-sh]

It is geometry. The direction ##\vec{k}## defines an orthogonal plane of equal momentum.

Why we write the expression of the plane wave like this [itex]\psi(x,t)[/itex]=Acos(kx-wt)? What is the problem If we write it like this Acos(kx+wt)?

 

[blue_leaf77]

Why we write the expression of the plane wave like this [itex]\psi(x,t)[/itex]=Acos(kx-wt)? What is the problem If we write it like this Acos(kx+wt)?

Note that by switching from ##e^{i(kx-\omega t)}## to ##\cos(kx-\omega t)##, you are also switching the discussion from that of quantum mechanical wavefunction to that of a classical wave because the latter is not a solution of the time dependent Schroedinger equation for free space.
The function ##\psi(x,t)=Ae^{i(kx-\omega t)}## is called a plane wave solution because for any time ##t##, the surface of constant phase, ##kx-\omega t =C##, is a plane perpendicular to the x axis.

 

[amjad-sh]

What about the other parts of my question?